Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $20.6$ years; the standard deviation is $1.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living between $19.5$ and $23.9$ years.
Solution: $20.6$ $19.5$ $21.7$ $18.4$ $22.8$ $17.3$ $23.9$ $99.7\%$ $68\%$ $15.85\%$ $15.85\%$ We know the lifespans are normally distributed with an average lifespan of $20.6$ years. We know the standard deviation is $1.1$ years, so one standard deviation below the mean is $19.5$ years and one standard deviation above the mean is $21.7$ years. Two standard deviations below the mean is $18.4$ years and two standard deviations above the mean is $22.8$ years. Three standard deviations below the mean is $17.3$ years and three standard deviations above the mean is $23.9$ years. We are interested in the probability of a porcupine living between $19.5$ and $23.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $68\%$ of the porcupines will have lifespans within 1 standard deviation of the mean. The probability of a particular porcupine living between $19.5$ and $23.9$ years is ${68\%} + \color{orange}{15.85\%}$, or $83.85\%$.